# Ex.5.4 Q1 Arithmetic progressions Solutions - NCERT Maths Class 10

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## Question

Which term of the A.P. $$121, 117, 113 ...$$is its first negative term?

[Hint: Find $$n$$ for $${a_n} < 0$$]

Video Solution
Arithmetic Progressions
Ex 5.4 | Question 1

## Text Solution

What is Known?

A.P. Series: $$121,\, 117, \,113 ...$$

What is Unknown?

First negative term of the AP

Reasoning:

$$n\rm{th}$$ term of an AP is $$\,{a_n} = a + \left( {n - 1} \right)d.$$

Where $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

Given:

• First Term, $$a = 121$$
• Common difference, $$d = 117 - 121 = - 4$$

We know that $$n^\rm{th}$$ term of AP, $${a_n} = a + \left( {n - 1} \right)d$$

\begin{align}{a_n} &= 121 + \left( {n - 1} \right) \times \left( { - 4} \right)\\ &= 121 - 4n + 4\\ &= 125 - 4n\end{align}

For the first negative term of this A.P,

\begin{align}{a_n} &< 0\\125 - 4n &< 0\\125 &< 4n\\n &> \frac{{125}}{4}\\n &> 31.25\end{align}

Therefore, $$32\rm{nd}$$ term will be the first negative term of this A.P.

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